Given, that `tanx=-5/12`, and x lies in second quadrant.
`therefore tanx=-5/12 rArr cotx=1/(tanx)`
`rArr cotx=1/(-5/12) = 1 xx -12/5 = -12/5`
`(therefore` x lies in second quadrant, therefore, sin and "cosec" ratio willl be positive and the remaining ratio tan, cot and sec will be negative.) then `sec^(2)x=-13/12`
`therefore cos=1/(secx)`
`rArr cosx=1/(-13/12) = 1 xx (-12/13)`
`rArr cosx=-12/13`
`therefore "cosec"^(2)x=cot^(2)x`
`therefore "cosec"^(2)x=1+(-12/5)^(2)= 1+144/25=169/25=(13/5)^(2)`
`rArr "cosec"x=13/5`
`therefore sinx=1/("cosec"x) rArr sinx=1/(13/5) = 1 xx 5/13`
`therefore sinx=5/13`
Therefore, `sinx=5/13`, `"cosec"x=13/5, cosx=-12/13`
`cotx=-12/5, secx=-13/12`. Ans.
